Math Problem Statement

give a simple solution for all and with understandable explanation

give a simple solution for all and with understandable explanation

Solution

Problem Breakdown and Solution:

The function given is f:RRf: \mathbb{R} \to \mathbb{R}, where f(x)=sinxf(x) = \sin x.

We need to answer two questions:

  1. (i) Determine the image set of ff under R\mathbb{R} and conclude whether ff is surjective.
  2. (ii) Solve for xx such that f(x)=12f(x) = \frac{1}{2} and conclude whether ff is injective.

Part (i): Image Set and Surjectivity

Step 1: Behavior of the sine function

The function sinx\sin x has the following properties:

  • The range (or image) of sinx\sin x is [1,1][-1, 1] because the sine function only takes values in this interval.
  • Even though xRx \in \mathbb{R} (all real numbers), the output of sinx\sin x is always between 1-1 and 11.

Step 2: Surjectivity

A function f:ABf: A \to B is surjective (onto) if every element in BB (codomain) has a preimage in AA.

Here:

  • The codomain is R\mathbb{R} (all real numbers).
  • The range (actual output values) of sinx\sin x is only [1,1][-1, 1].

Since the codomain R\mathbb{R} includes all real numbers, but f(x)f(x) does not cover values outside [1,1][-1, 1], the function ff is not surjective.


Part (ii): Solving f(x)=12f(x) = \frac{1}{2} and Injectivity

Step 1: Solving f(x)=12f(x) = \frac{1}{2}

The equation sinx=12\sin x = \frac{1}{2} can be solved as follows:

  • The sine function equals 12\frac{1}{2} at specific angles: x=π6+2nπandx=ππ6+2nπfor nZ.x = \frac{\pi}{6} + 2n\pi \quad \text{and} \quad x = \pi - \frac{\pi}{6} + 2n\pi \quad \text{for} \ n \in \mathbb{Z}.
  • This simplifies to: x=π6+2nπandx=5π6+2nπfor nZ.x = \frac{\pi}{6} + 2n\pi \quad \text{and} \quad x = \frac{5\pi}{6} + 2n\pi \quad \text{for} \ n \in \mathbb{Z}. Here, nn represents all integers because the sine function is periodic with period 2π2\pi.

Step 2: Injectivity

A function ff is injective (one-to-one) if different inputs x1x_1 and x2x_2 result in different outputs: f(x1)=f(x2)    x1=x2f(x_1) = f(x_2) \implies x_1 = x_2.

Here:

  • The sine function is periodic, meaning it repeats its values every 2π2\pi.
  • For f(x)=12f(x) = \frac{1}{2}, there are infinitely many xx values (e.g., π6\frac{\pi}{6}, 5π6\frac{5\pi}{6}, and their periodic extensions).

Since multiple inputs give the same output, ff is not injective.


Final Answers:

  1. The image set of ff is [1,1][-1, 1], and ff is not surjective.
  2. The equation f(x)=12f(x) = \frac{1}{2} has multiple solutions, so ff is not injective.

Clarification Questions or Follow-ups:

  1. What is the domain and range of trigonometric functions?
  2. Why is the sine function periodic?
  3. How do you determine surjectivity and injectivity?
  4. Can a function be injective but not surjective?
  5. What are the general solutions for sinx=k\sin x = k?

Tip: To check surjectivity and injectivity, always compare the range of the function with its codomain and analyze for repeated inputs.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Functions
Surjectivity
Injectivity
Trigonometry
Sine Function

Formulas

f(x) = sin(x)
General solutions for sin(x) = k

Theorems

Definition of surjective functions
Definition of injective functions
Periodicity of sine function

Suitable Grade Level

Grades 11-12